Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{-4}{4z + 24} \div \dfrac{2z}{3z(z + 6)} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{-4}{4z + 24} \times \dfrac{3z(z + 6)}{2z} $ When multiplying fractions, we multiply the numerators and the denominators. $t = \dfrac{ -4 \times 3z(z + 6) } { (4z + 24) \times 2z } $ $ t = \dfrac {-4 \times 3z(z + 6)} {2z \times 4(z + 6)} $ $ t = \dfrac{-12z(z + 6)}{8z(z + 6)} $ We can cancel the $z + 6$ so long as $z + 6 \neq 0$ Therefore $z \neq -6$ $t = \dfrac{-12z \cancel{(z + 6})}{8z \cancel{(z + 6)}} = -\dfrac{12z}{8z} = -\dfrac{3}{2} $